Difference between revisions of "2012 AIME I Problems/Problem 2"

Latest revision as of 17:30, 20 January 2024

Problem

The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence.

Solutions

Solution 1

If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195}.$

Alternatively, notice that in the original sequence, $11a_1 + 55d = 715$ , from which $a_1 + 5d = 65$ . Since we are tasked to find $a_1 + a_6 + a_{11} = 3(a_1 + 5d)$ , the desired answer is $3 \cdot 65 = \boxed{195}.$

Solution 2

After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered around the mean, our final answer is $65 \cdot 3 = \boxed{195}$

Solution 3

Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \boxed{195}$ .

~ cxsmi

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/298

~ dolphin7

Video Solution

https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=4689

~ pi_is_3.14

@@ Line 13: / Line 13: @@
 ===Solution 3===
-Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, it can be noted that <math>\frac{2a_1 + 10d}{2} \cdot 11 = 715</math> or <math>2a_1 + 10d = 130</math> for all sets of first terms and common differences that fit the conditions given in the problem. Without loss of generality, suppose that <math>a_1 = 60</math> and <math>d = 1</math>. Then the first term of the corresponding arithmetic sequence will be <math>60</math>, the sixth (middle) term will be <math>65</math>, and the eleventh (largest) term will be <math>70</math>. Thus, the final answer is <math>60 + 65 + 70 = \boxed{195}</math>.
+Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that <math>\frac{2a_1 + 10d}{2} \cdot 11 = 715</math> or <math>2a_1 + 10d = 130</math> for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that <math>a_1 = 60</math> and <math>d = 1</math>. Then the first term of the corresponding arithmetic sequence will be <math>60</math>, the sixth (middle) term will be <math>65</math>, and the eleventh (largest) term will be <math>70</math>. Thus, our final answer is <math>60 + 65 + 70 = \boxed{195}</math>.
+~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 == Video Solution by Richard Rusczyk ==

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search